Capacitance in Series: Questions for David

[meldave00]

have a question in regards to capacitors in series. We are taught that the charge must be equal on all plates of the capacitors in the case of a series connection. However, we also know that Q = CV. If we increase V, then Q must increase. How is this possible for the node between the two capacitors. Does it not have a finite amount of charge? Also, does not all the charge at the middle node just immediately rush to the perspective plates no matter what voltage is applied? Or does just some charge rush to the plates?

Hope this isn't too confusing. I will clarify more if people respond.

regards,

David

 

[Gokul43201]

You'll have to explain what you mean by "node".

What happens with a typical capacitor is that when you apply a potential difference (connect a battery) across the plates (or electrodes) of a capacitor you are providing a path and a reason for charge to travel from one plate to the other. The total charge is conserved - all that happens is a kind of rearrangement that minimizes energy (or balances forces).

 

[benzun_1999]

Your resoning might seem logical but not scientific though Q=CV , the charge being withdrawn from the battery is due to the net capacitance due to the entire series combination. Now the charge on all plates remain same due to conservation of charge. So even if u change the capacitance of a capactior this equality remains ie the charge on all the plates will be same.

-Benzun

 

[meldave00]

Let me reiterate.

Imagine two caps in series and a voltage source connected across both of them. In order to calculate the individual voltage drops across each cap you need to use the Q=CV relationship for each. The first thing that you need to do is assume that the charge Q is the same on both caps in order to solve for the individual voltage drop across each cap. I made this assumption and was able to solve the voltage drops across each cap correctly. However, it got me thinkin. How does the charge on the connection between the caps increase. If the caps remain the same value, but you just increase V then the charge Q must increase as well.

But how does it increase on the conductor between the caps??

regards

David

 

[Gokul43201]

Let me reiterate.

Imagine two caps in series and a voltage source connected across both of them. In order to calculate the individual voltage drops across each cap you need to use the Q=CV relationship for each. The first thing that you need to do is assume that the charge Q is the same on both caps in order to solve for the individual voltage drop across each cap. I made this assumption and was able to solve the voltage drops across each cap correctly. However, it got me thinkin. How does the charge on the connection between the caps increase. If the caps remain the same value, but you just increase V then the charge Q must increase as well.

It does not. What increases is the absolute value of the charge. To reiterate what I said above, increasing the voltage only takes more charge from cap#1 and transfers it across the node to cap#2 . As a result this plate of cap#1 becomes more negatively charged and the plate on cap#2 (that it is connected to) becomes more positively charged. But the total charge is still zero.

Remember that with the equation you wrote down, Q does not represent the total charge in the capacitor. It represents the absolute value of the charge on either of the electrodes of the capacitor. If one electrode has charge Q, the other has charge -Q.

 

[meldave00]

First of all. Thank you for the reply. You have no idea how much I have been thinking of this problem.

Second. Please bare with me during this discussion. Please be patient with me on this one. I may take me awhile to understand this one.

So. Let me get a couple of things clear.
1). Are you saying that this is a false statement?
abs(Q1pos) = abs(Q1neg) = abs(Q2pos) = abs(Q2neg), where pos represent the top plates charge and neg represent the bottom plates charge.

2). I'm assuming that negative charge is electrons and positive charge is lack of electrons?

3). If #2 is true. Does not just all the electrons move to the bottom plate of Cap#1 with the application of any voltage across the caps? If the supply voltage increases, how can more electrons move towards the bottom plate of cap#1 from top plate of cap#2. The top plate of cap#2 does not have any more electrons to "transfer" as you call it.

 

[quantumdude]

But how does it increase on the conductor between the caps??

The charge doesn't increase on the node between the caps. I think you're missing the idea of displacement current here. You have a single conduction current [itex]i[/itex] that enters the first cap. That conduction current stops dead at the positive plate as positive charge builds up there (it's really negative charges leaving, but whatever). That conduction current doesn't flow across the capacitor plates, but rather is replaced by the displacement current [itex]i_d=\epsilon d\Phi/dt[/itex], which is exactly equal to the conduction current. Thus, the current is continuous.

On the opposite plate of the same capacitor (the plate that is connected to the node you speak of), negative charge builds up and pushes the positive charge away from it. Thus, the conduction current resumes. Observe that there is no charge stored here: the charge flows until it hits the near plate of the other capacitor, at which point another displacement current builds up, and the above process repeats itself.

 

[nrqed]

Let me reiterate.

Imagine two caps in series and a voltage source connected across both of them. In order to calculate the individual voltage drops across each cap you need to use the Q=CV relationship for each. The first thing that you need to do is assume that the charge Q is the same on both caps in order to solve for the individual voltage drop across each cap. I made this assumption and was able to solve the voltage drops across each cap correctly. However, it got me thinkin. How does the charge on the connection between the caps increase. If the caps remain the same value, but you just increase V then the charge Q must increase as well.

But how does it increase on the conductor between the caps??

regards

David

By cap I guess you mean what I would call a plate of the capacitor. I don't know what you mean by the "value of the cap". I guess you mean the capacitance, which depends on the size of the plates, the distance between them and the material between the plates (for parallel plate capacitors, there are also a lot of different types of capacitors).

Let's sya there are two parallel plate capacitors (with air between the plates) in series.

You have

wire - plate - air - plate - wire - plate - air - plate -wire


And you connect this to a battery. If you increase the potential difference applied to the batteries, and you focus on the wire conencting the two capacitors, what happens is that some positive charges will move to one of the plates and and equal amount of negative charge (in absolute value) will flow to the other plate. The net charge of the wire conencting the two capacitors does not change (it remains zero). The distribution of the charges has changed. There is more charge on the two capacitors but if you look at the polarities you will see (say)

positive plate-air-negative plate-wire-positive plate-air-negative plate

That is, the negative plate (cap) of one capacitor is connected to the positive plate of the next capacitor and so on if there are several capacitors.

I am not sure if that adresses your question. I hope.

 

[meldave00]

Capacitor in Series

Again thank you for your patience in this matter. I hope you stick with me on this matter. I'm getting close to understanding. By the way, what is your name or handle? (Don't have to give it to me if you don't wish)

I like your word picture of the capacitor circuit. I've added GND and Vsupply to complete the circuit. I've apppended numbers for each plate and wire. I will be referring to each plate/wire/air numbers throughout the message.

GND - wire1 - plate1 - air1 - plate2 - wire2 - plate3 - air2 - plate4 -wire3 - Vsupply

where : plate1 and plate2 make capacitor1
and plate3 and plate4 make capacitor2

Here is what I understand (Stop me if I'm incorrect):
1). When Vsupply is initially applied to this circuit. Electrons will build up on plate1 which in turn repels electrons away from plate 2 thus leaving positive charge built up on plate2. The electrons leaving plate2 causes electron to build up on plate3 which in turn repels electrons away from plate4(leaving behind positive charge) and the process is done and the circuit settles into equilibrium when Vcapacitor1+Vcapacitor2 = Vsupply. There is no current flowing thru the capacitors per se, but just displacement currents when Vsupply is first applied. No currents or displacement currents flow thru the circuit after the circuit settles into equilibrium.

2). Quote from my physics book. "For a series combinations of capacitors, the magnitude of the charge must be the same on all the plates when in equilibrium." That is Q-(plate1) = Q+(plate2) = Q-(plate3) = Q+(plate4). By the way, you can look this up in any physics book that explains two capacitors in series.

Here are my questions again:
It's obvious that I'm interested in the region that consists of plate2 - wire2-plate 3

1). When VSupply is increased then the voltage across each capacitor will increase which in turn will increase all the charge on each capacitor according to the Q = CV equation where Q = Q-(plate1) = Q+(plate2) = Q-(plate3) = Q+(plate4). This must mean that more additional electrons must leave plate2 and migrate over to plate3? My question is: How can there be more electrons to move from plate 2 onto plate3 with the addition of more voltage. Why doesn't "all" the electrons move over to plate3 with any small amount of Vsupply. Since the plate2 - wire2-plate 3region is a conductor (very low resistance). I would think that all the electrons would flow to plate 3 with any small amount of Vsupply voltage and then not change with any additional Vsupply voltage because "all" the electrons would have already migrated over to plate3.

1a) According to Gauss's Law, there is no "net" charge stored inside a conductor when in equilibrium and there is no electric field inside a conductor when in equilibrium. Also, all "net" charge must reside on the surface of a conductor when in equilibrium. So in the case of plate2 - wire2-plate3 region (which is a conductor), plate3 must borrow more additional electrons when Vsupply is increased. Is it borrowing the additional electrons from the surface of plate2 or from within the wire2? In which case if it is borrowing more electrons from wire2 then the electrons would not be considered "net" charge, but some other type of charge. Is there a difference? You stated earlier that the "net" charge within the plate2 - wire2-plate3 region is zero.

2). You stated that the net charge is zero in the plate2 - wire2-plate3 region. However, there is negative charge stored on plate3 and positive charge stored on plate2. I am wondering why this does not create potential difference form plate2 to plate 3? I know that it does not because its a conductive wire in the circuit that has very low resistance. However, I was under that assumption that charge imbalance would create an Efield which would create a voltage drop. Obviously I am wrong, but can you explain?

3). Last but not least. This is a doozy!. What is voltage? I know that voltage is a scalar quantity the equals the amount of work done to move a positive point charge from A to B within an electric field. It the line integral of the efield across some distance ds. This is the physics book definition, but what does it mean from a charge perspective. Say you have a node A in a circuit that is a very long distance away from node B. The nodes are far away from each other as to not be effected by any Efield produced by the charges that resides in the nodes. Say NodeA is 5V and NodeB is 3 Volts. What is the difference between the nodes from a charge standpoint? Is there more electrons in NodeB than NodeA? Or is there a greater charge density in NodeB than NodeA?

 

[nrqed]

Again thank you for your patience in this matter. I hope you stick with me on this matter. I'm getting close to understanding. By the way, what is your name or handle? (Don't have to give it to me if you don't wish)

Hi .Myname is Patrick

I like your word picture of the capacitor circuit. I've added GND and Vsupply to complete the circuit. I've apppended numbers for each plate and wire. I will be referring to each plate/wire/air numbers throughout the message.

GND - wire1 - plate1 - air1 - plate2 - wire2 - plate3 - air2 - plate4 -wire3 - Vsupply

where : plate1 and plate2 make capacitor1
and plate3 and plate4 make capacitor2

Ok

Here is what I understand (Stop me if I'm incorrect):
1). When Vsupply is initially applied to this circuit. Electrons will build up on plate1 which in turn repels electrons away from plate 2 thus leaving positive charge built up on plate2. The electrons leaving plate2 causes electron to build up on plate3 which in turn repels electrons away from plate4(leaving behind positive charge) and the process is done and the circuit settles into equilibrium when Vcapacitor1+Vcapacitor2 = Vsupply. There is no current flowing thru the capacitors per se, but just displacement currents when Vsupply is first applied. No currents or displacement currents flow thru the circuit after the circuit settles into equilibrium.

Right

2). Quote from my physics book. "For a series combinations of capacitors, the magnitude of the charge must be the same on all the plates when in equilibrium." That is Q-(plate1) = Q+(plate2) = Q-(plate3) = Q+(plate4). By the way, you can look this up in any physics book that explains two capacitors in series.

Right

Here are my questions again:
It's obvious that I'm interested in the region that consists of plate2 - wire2-plate 3

1). When VSupply is increased then the voltage across each capacitor will increase which in turn will increase all the charge on each capacitor according to the Q = CV equation where Q = Q-(plate1) = Q+(plate2) = Q-(plate3) = Q+(plate4). This must mean that more additional electrons must leave plate2 and migrate over to plate3? My question is: How can there be more electrons to move from plate 2 onto plate3 with the addition of more voltage. Why doesn't "all" the electrons move over to plate3 with any small amount of Vsupply. Since the plate2 - wire2-plate 3region is a conductor (very low resistance). I would think that all the electrons would flow to plate 3 with any small amount of Vsupply voltage and then not change with any additional Vsupply voltage because "all" the electrons would have already migrated over to plate3.

As the electrons move away from plate 2, plate 2 becomes more and more positive obviously. That means that it gets harder and harder to move electrons away from plate 2 (more positively charged plate 2 is, the more the electrons are attracted by it!).

The key point is that there are two effects: the electrons of plate 1 are repelling the electrons of plate 2, pushing them toward plate 3. But at the same time, thenet positive charge of plate 2 are *attracting* electrosn toward it. Equilibrium is reached when the two effects cancel (that depends on how much charge is accumulated on plate 1, which depends on the capacitance of the first capacitor and on how "strong" the battery is.)

1a) According to Gauss's Law, there is no "net" charge stored inside a conductor when in equilibrium and there is no electric field inside a conductor when in equilibrium. Also, all "net" charge must reside on the surface of a conductor when in equilibrium. So in the case of plate2 - wire2-plate3 region (which is a conductor), plate3 must borrow more additional electrons when Vsupply is increased. Is it borrowing the additional electrons from the surface of plate2 or from within the wire2?

Well, it would be hard to follow individual electrons so it's hard to say But what is most likely to be happening is that electrons are pushed away from plate 2 and these push away some nearby electrons which then push away nearby electrons and so on in a domino-type effect.

In which case if it is borrowing more electrons from wire2 then the electrons would not be considered "net" charge, but some other type of charge. Is there a difference? You stated earlier that the "net" charge within the plate2 - wire2-plate3 region is zero.

The net charge is simply= (e times the number of protons - e times the number of electrons). No net charge means that the number of electrons is equal to the number of protons. For an isolate system the net charge can never change. The net charge of the system plate 2+ wire 2+ plate 3 does not change.

2). You stated that the net charge is zero in the plate2 - wire2-plate3 region. However, there is negative charge stored on plate3 and positive charge stored on plate2. I am wondering why this does not create potential difference form plate2 to plate 3? I know that it does not because its a conductive wire in the circuit that has very low resistance. However, I was under that assumption that charge imbalance would create an Efield which would create a voltage drop. Obviously I am wrong, but can you explain?

There is also an E field created by the charges on plate 1. It's the total E field which cancels out, at equilibrium.


More in a minute

Patrick

 

[nrqed]

3). Last but not least. This is a doozy!. What is voltage? I know that voltage is a scalar quantity the equals the amount of work done to move a positive point charge from A to B within an electric field. It the line integral of the efield across some distance ds. This is the physics book definition, but what does it mean from a charge perspective. Say you have a node A in a circuit that is a very long distance away from node B. The nodes are far away from each other as to not be effected by any Efield produced by the charges that resides in the nodes. Say NodeA is 5V and NodeB is 3 Volts. What is the difference between the nodes from a charge standpoint? Is there more electrons in NodeB than NodeA? Or is there a greater charge density in NodeB than NodeA?

It has nothing to do with the amount of charge at the nodes or with charge density. And notice that there is no accumulation of charges at nodes (unless by nodes you would mean the plates of a capacitor). The charges are constantly flowing.

The electric potential is not a measure of how much charge is at a point.
The best analogy is in terms of gravity. Imagine skiers in a ski resort, taking a ski lift, going uphill , skiing down etc. Then the potential at a point is the analogue of the "height" (or altitude if you will) of a point.The larger the potential difference between two points is, the more work is required to move a charge between those two points (if it is on the way "up") or the more energy it will give away between those two points (if it moves on the way "down").

(here up means in the direction of increasing potential if the charge is positive and in the direction of decreasing potential if the charge is negative. A negative charge is like a "negative mass" in this analogy)

 

[meldave00]

Capacitor in Series

So in the plate2 - wire2 - plate3 region there are multiple efields occurring.
One from plate 2 to plate 1 and one from plate2 into the wire 2 toward plate 3. These two Efield cancel out so that there is no potential drop across theplate2 - wire2 - plate3 region.

So there is either a step boundary of positive charge to negative charge going from plate2 toward wire2 and plate3. Or there has to be a gradual gradient of electrons that are at a minimum closest to plate2 and a maximum on plate3? Any idea?

I'm invisioning that all the electrons are not accumulated on plate 3, but is spread out across the wire2 to plate 3 connection where the electrons are maximum on plate 3.

Also, for the Voltage question. I've heard the analogy for uphill mountain potential comparision for voltage. However, no matter where you are on the hill there is still a constant force being exerted on an object no matter where they are on the hill.

However, in a electric circuit. I believe that it is different. I believe that it is more of a water analogy. The pressure of water is analogous to voltage. The greater the pressure the greater the voltage. This means that that there is more force on the electrons at the lower potential V and less force being exerted on the electrons at higher potential V. This is different than a hill analogy because an object on a hill always has the same amount of force on it no matter where its at on the hill. Granted the higher the object is the more work it can do. But still if has the same force of gravity on it no matter what.

 

[meldave00]

Capacitors in Series

Sorry for the multiple threads. I'm new to this forum. I'll be better from here on out. I will consider this thread the main thread. Again sorry for the multiple threads. I don't want to scare people away.

 

[nrqed]

Also, for the Voltage question. I've heard the analogy for uphill mountain potential comparision for voltage. However, no matter where you are on the hill there is still a constant force being exerted on an object no matter where they are on the hill.

That is not true. The force depends on how steep the slope is. The slope is not necessarily of equal steepness everywhere! This allows for a varying force. There may even be sections which are flat so that the force is zero (here by force I obviously mean the net force)

However, in a electric circuit. I believe that it is different. I believe that it is more of a water analogy. The pressure of water is analogous to voltage. The greater the pressure the greater the voltage. This means that that there is more force on the electrons at the lower potential V and less force being exerted on the electrons at higher potential V.

But it is NOT true that the lower the potential, the stronger the force on an electron is!

An electron could be at a point where the potential is one gazillion volts and feel no force at all! Another electron could be at a point where the potential is zero and feel a huge force!

One way to see that the value of the electric potential at one point is unrelated to the force is that the value of the potential at any given point is arbitrary. By changing the choice of ground, the value of the potential will change. But of course the value of the force can't change.
(in a circuit, the potential at a given node can be defined to be anything you want, for example).

What *does* determine the force is the rate of change of the potential with respect to the position. So the potential is really analogue to the *height* of a point on a hill (you can put the origin of your coordinate system at the top of the hill, 100 m above the top of the hill, at the center of the Earth or, as people usually do, at sea level so the height could be anything). If you want to determine the force on a ball placed at some point on the hill, just knowing the height of that point is useless. what you need to know is the gradient of the height at that point, basically how steep the hill is. It's the same thing for the electric potential. You eed to know how it *varies* near the point where your charge is placed.

 

[meldave00]

Capactitor in Series

Nrqed,

I started a thread in "Electrical Engineering" under the subject of "Voltage". I'm going to reply to your response there. I felt the subject of voltage deserved its own thread. m Also, please read the history of replies prior to my response to you.

regards,

David

 

[Firefox123]

Hi David...

I had typed a reply to one of your posts in the other thread about capacitors, but I was away for the week and never posted it.

Since this appears to be the active thread I will post it here. Hopefully you will find something useful in my reply.

2). You stated that the net charge is zero in the plate2 - wire2-plate3 region. However, there is negative charge stored on plate3 and positive charge stored on plate2. I am wondering why this does not create potential difference form plate2 to plate 3? I know that it does not because its a conductive wire in the circuit that has very low resistance. However, I was under that assumption that charge imbalance would create an Efield which would create a voltage drop. Obviously I am wrong, but can you explain?

Keep in mind that plate3 is negative with respect to other specific points or nodes in the circuit and the same applies to plate2. We put a '+' symbol on plate2 because we are denoting a voltage drop over plate2-air1-plate1. We put a '-' symbol on plate3 because we are denoting a voltage drop over plate4-air2-plate3.

Think about it like this...the plate3/plate2 “node” is "more negative" than plate4, but "less negative" than plate1.

3). Say you have a node A in a circuit that is a very long distance away from node B. The nodes are far away from each other as to not be effected by any Efield produced by the charges that resides in the nodes. Say NodeA is 5V and NodeB is 3 Volts. What is the difference between the nodes from a charge standpoint? Is there more electrons in NodeB than NodeA? Or is there a greater charge density in NodeB than NodeA?

Keep in mind that when you say “NodeA is 5 V and NodeB is 3 Volts” you have to have some kind of a point of reference…..Node A is 5 Volts and Node B is 3 volts with respect to what? Let's say we have a third “node”…Node ‘C’ which is ground.

What are we really saying? We are saying that Node A is 2 volts higher in potential (which is potential energy per unit charge) than Node B is and if we put a “test” charge in the potential field of Node A and Node B we would notice a difference of 2 Volts in the potential energy per unit charge.



Russ

 

[meldave00]

Russ,

Thanks for the reply. By the way, how do you quote other peoples words?
I haven't figured that out yet.

Any in response to your response. You said "Think about it like this...the plate3/plate2 “node” is "more negative" than plate4, but "less negative" than plate1."

What I'm saying is that there is a charge inbalance in the node between the two capacitors. So why is there not a voltage drop? I know that the E-Field inside a conductor is zero when in equilibrium.

regards,

David

 

[Firefox123]

Russ,

Thanks for the reply. By the way, how do you quote other peoples words?
I haven't figured that out yet.

Ill put an extra character that you would not include in the example by using a '*'

To quote someone type...[*quote=whoeveryouarequoting]*Insert the persons text here*[*/quote]So the commands are [operation]thing to operate on [/operate] where the '[]' signify the operation and the '/' is the close for that operation.

To make text bold type [*b]*Insert text here[*/b]

Any in response to your response. You said "Think about it like this...the plate3/plate2 “node” is "more negative" than plate4, but "less negative" than plate1."

What I'm saying is that there is a charge inbalance in the node between the two capacitors. So why is there not a voltage drop? I know that the E-Field inside a conductor is zero when in equilibrium.

Okay...Im going to give this a shot. Someone please correct me if I am wrong.

Here is what I believe happens...

Lets focus on the node in between the two capacitors. So we will look at both plates plus the wire conductor in between.

When the caps are charging we have one plate that basically deposits electrons onto the other plate via the wire conductor.

We still have the same number of charges in the plate-wire-plate system, but the charges are now distributed differently than they were before.

On one plate we have a '+' charge because electrons were removed and on the other plate we have a '-' charge because electrons were "added".

But what about the wire in between? I would imagine that the free electrons in the wire are not evenly distributed along the wire but are displaced the exact opposite of the plates.

In other words, I would assume a higher electron density near the positive plate than near the negative plate. So an opposite electric field is created in the wire so the net field is zero.

Some might object at this point and ask why the electrons don't flow onto the positive plate and replace the charge deficiency there...I would answer that the electric field across the capacitor prevents this from occurring on both capacitors.

So we have four different fields in this situation.

Fields #1 and #2: The two fields across both capacitors repel and attract charges to and from the plates and create a potential energy between the plates. If we put a test charge in this region, it will flow from one point to another.

Fields #3 and #4: These two fields are set up by the charge distributions in between the two capacitors.

Field #3 is the field between the plates caused by the local distribution of extra electrons on one plate and missing electrons on the other plate.

Field #4 is the field created by a non-uniform distribution of charge in the wire conductor and cancels out field 3 such that a test charge placed anywhere in this region will experience no net electric field acting on it.So my short answer is that the reason there is no voltage drop is because the charge distribution in the wire cancels out the electric field created by the charge imbalance. Since there is then no net electric field, there is no voltage drop.
I hope that helps David...this is the best I can do for a quick answer. Someone please correct me if I am way off here.
Russ

****edited to add****

A thought just occurred to me about this...

Perhaps it is not the charge distribution in the wire, but the charge distribution on the plates themselves that causes this opposite electric field.

Think about a single capacitor that is charged ...as one side becomes "negative" the other becomes "positive"...so the "positive" charged plate helps to hold the electrons in place on the "negative" charged plate.

Now disconnect the capacitor and think about the molecules and free electrons on both plates...

The negative plate has "extra" electrons on its surface that repel the remaining free electrons on the positive plate to the outside of the plate.

So if we looked at charge distribution from left to right going from negative plate to positive plate we get... '-' '+' '-'

Perhaps this distribution is what creates the opposing electric field bewteen the two capacitors.As a final thought maybe we can say that each individual situation is different depending on the value of the capacitors and how close they are together, but an opposing field is set up within the plate-wire-plate interface by the distribution of charges on the capacitor plates combined with the effect this has on the free electrons in the wire.

Does that make sense?

I would be surprised if the electron distribution in the wire in between the capacitors is uniform for all cases of two series capacitors.

Russ

 

[meldave00]

Russ,

I like what you are saying. I need to think about what you said a little longer.

But in the mean time here is something for you to think about. Gauss's Law keep poppin in my head when I think about your response. Gauss's Law says that all excess charge lies on the surface of a conductor in equilibrium and that there is no excess charge inside a conductor in equilibrium and the net E field inside a conductor is zero in a conductor in equilibrium. I'm wondering if I'm misintrepreting this law incorrectly or not. How does Gauss's Law stay valid for the case of the node between two capacitors in series. Again, I'm still thinking about your answer. You might be on to something.

bye for now,

David

 

[Firefox123]

Hello again Dave...

Russ,

I like what you are saying. I need to think about what you said a little longer.

Take your time...

But in the mean time here is something for you to think about. Gauss's Law keep poppin in my head when I think about your response. Gauss's Law says that all excess charge lies on the surface of a conductor in equilibrium and that there is no excess charge inside a conductor in equilibrium and the net E field inside a conductor is zero in a conductor in equilibrium. I'm wondering if I'm misintrepreting this law incorrectly or not. How does Gauss's Law stay valid for the case of the node between two capacitors in series. Again, I'm still thinking about your answer. You might be on to something.

First the quick answer...I would possibly rephrase your interpretations slightly

It is true that

1. All excess charge lies on the surface of a conductor in equilibrium.

2.There is no excess charge inside a conductor in equilibrium.

3.The net E field inside a conductor is zero in a conductor in equilibrium.

Assuming that anyone of these 3 statements are true, we can then use Guass's law to prove the other two.


Okay...Guass's relates the flux of an electrostatic field over some surface to the charge enclosed by that surface.

So Guass's law looks like this:
[tex]\iint_{\mathcal{S}}\vec{E}(x,y,z)\cdot{\vec{n}} dS = \frac{q} {\epsilon}[/tex]

where [tex]q[/tex] is the enclosed charge and [tex]\epsilon[/tex] is the permittivity of free space.

So the "flux" or "amount of flow" (not to be taken literally, since the electric field really doesn't "flow" like a fluid does) of the electric field through the surface is
[tex]\iint_{\mathcal{S}}\vec{E}(x,y,z)\cdot{\vec{n}} dS[/tex]


Now to answer your question...

Suppose we have some conductor and we put a bunch of electrons inside the conductor...what will happen?

The electrons will repel each other and move as far apart as possible which means they will be on the outside of the conductor. So we have a bunch of free electrons just sitting on the outside of the conductor.

So why would the electric field be zero inside the conductor even though each electron is surrounded by an electric field?

The reason why it is zero is the same reason why the charges eventually stop moving...because a state of equilibrium has been reached. Each charge experiences a net force of zero from the other charges because they are distrubuted in such a way that the fields end up cancelling each other out.

Kind of like one person pulling your right arm while another pulls your left arm with equal force.

The same is true for the plate-wire-plate region. We have a case of equilibrium.

Picture the two capacitors plus the wire in bewteen...go from left to right with the negative side of the battery on the left..Ive exagerrated the spacing between the capacitor plates to make it easier to visualize. The charge signs in () are the induced charges on the outer edge of the plate away from the opposite capacitor plate.

CAP 1 ***wire in between*** CAP 2
- +(-) ______________________ - +(-)

Before equilibrium is reached the negative plate of CAP 1 (connected to the negative plate of the battery) forces electrons to leave the positive plate of CAP 1 and go onto the negative plate of CAP 2

As charge builds up on CAP 2 it increasingly resists this push from CAP 1 until equilibrium is reached and the "push" from CAP 2 equals the "push" from CAP 1.

I hope that makes sense.


***Curiousity question...what would happen is we removed the wire along with the two plates connected to it?


I would think that the part of the circuit still connected to the battery would not change but would just be an open circuit at battery potential...but what about the section no longer connected to the circuit?

What about the plate-wire-plate section? I would imagine that the induced charge on the plate of CAP 1 would no longer be separated and the excess static charge on the plate of CAP 2 would flow through the wire to the CAP 1 plate as a static discharge...





Russ

 

[Firefox123]

Field #4 is the field created by a non-uniform distribution of charge in the wire conductor and cancels out field 3 such that a test charge placed anywhere in this region will experience no net electric field acting on it.

Correction...I think that statement should read...

"..such that the electrons anywhere in this region will experience no net electric field acting on it."

not "a test charge".

Introducing a test charge unnecessarily complicates the situation and I am not convinced that the introduced charge would experience no net electric field. It would likely move to the outside of the conductor and its location would be such that the new distribution of charges would once agian be in static equilibrium.

Minor correction, but there is no need for me to mention a test charge...we should just stick to thinking about the charges already there.

So the corrected comment is:

Field #4 is the field created by a non-uniform distribution of charge in the wire conductor and cancels out field 3 such that the electrons anywhere in this region will experience no net electric field acting on it.




Russ

 

[meldave00]

Russ,

I appreciate your responses. They seem to be well thought out. I need some time to carefully read thru your responses and formulate my own opinion. You will hear from me on this subject soon.

regards,

David

 

[Firefox123]

Russ,

I appreciate your responses. They seem to be well thought out. I need some time to carefully read thru your responses and formulate my own opinion. You will hear from me on this subject soon.

regards,

David

No problem. Hopefully I haven't confused you or given you incorrect answers.

Take your time to really think about this stuff.

If you think up or discover a better explanation let me know so I can know the correct answer also.


Later,

Russ

 

[meldave00]

Capacitors in Series

Russ,

I have been thinking a little more about your response. To me it makes sense unless anyone else has objections. I will adopt your theory as the way it is until someone proves otherwise.

I do have a couple of more questions or caveats in regards to your theory:

[*quote=steadele]*So we have four different fields in this situation.

Fields #1 and #2: The two fields across both capacitors repel and attract charges to and from the plates and create a potential energy between the plates. If we put a test charge in this region, it will flow from one point to another.

Fields #3 and #4: These two fields are set up by the charge distributions in between the two capacitors.

Field #3 is the field between the plates caused by the local distribution of extra electrons on one plate and missing electrons on the other plate.

Field #4 is the field created by a non-uniform distribution of charge in the wire conductor and cancels out field 3 such that a test charge placed anywhere in this region will experience no net electric field acting on it.
*[*/quote]

I'm a little confused on you definitions of field 3 and field 4

Here is my definition of E-Fields 1,2,3,4. Let me know if this fits your definition. My field numbers may differ from yours, but the there are still 4 main fields. Field 1 is due to the electrons on the neg. plate of CAP 1 pushing electrons away from the positive plate of CAP 1. This in turn creates field 3 which is the charge redistribution on the positive plate of CAP 1. This positive plate has negative charge on the wire side of the plate and positive charge on other side. Field 1 and Field 3 cancel each other out so the the net Efield in the positive plate of CAP 1 and wire connection is zero.

Field 2 is due to the electrons on the neg. plate of CAP 2 pushing electrons away from the positive plate of CAP 2. This in turn creates field 4 which is the charge redistribution on the positive plate of CAP 2. This positive plate has negative charge on the wire side of the plate and positive charge on other side. Field 2 and Field 4 cancel each other out so the the net Efield in the positive plate of CAP 2 and wire connection too supply is zero.

Before I go any further. Let me know if this fits your definition of what's going on. If it does than I have further questions for you.

regards,

David


CAP 1 ***wire in between*** CAP 2
- +(-) ______________________ - +(-)

 

[Firefox123]

Dave,,,

Quick note...Ill be replying to your post later on tonight.

I screwed up the definitions of fields #3 and #4...the situation is much simpler than my original description. I wasnt thinking about the plates correctly.

Field #4 is completely wrongi n my original definition and field #3 needs to be redefined. Ill give you what I think is a much simpler and much more accurate explanation later on tonight.

I feel so stupid for the mistake I made...Ill expalin later.

Till then.

Russ

 

[Firefox123]

Hi david...

Russ,

I have been thinking a little more about your response. To me it makes sense unless anyone else has objections. I will adopt your theory as the way it is until someone proves otherwise.

I do have a couple of more questions or caveats in regards to your theory:

After thinking about my responses I believe I need to correct a few errors in my answers...

I'm a little confused on you definitions of field 3 and field 4

Forget my initial definitions of those two fields...

I think the situation is much simpler than my "non uniform charge distribution" thought...I think I made a fundamental error in thinking about the wire in that manner and believe my statement about non unifrom charge distribution was incorrect...


As far as the fields go...the text I have highlighted in red is what I would remove from your definitions..

Here is my definition of E-Fields 1,2,3,4. Let me know if this fits your definition. My field numbers may differ from yours, but the there are still 4 main fields. Field 1 is due to the electrons on the neg. plate of CAP 1 pushing electrons away from the positive plate of CAP 1. This in turn creates field 3 which is the charge redistribution on the positive plate of CAP 1. This positive plate has negative charge on the wire side of the plate and positive charge on other side. Field 1 and Field 3 cancel each other out so the the net Efield in the positive plate of CAP 1 and wire connection is zero.

Field 2 is due to the electrons on the neg. plate of CAP 2 pushing electrons away from the positive plate of CAP 2. This in turn creates field 4 which is the charge redistribution on the positive plate of CAP 2. This positive plate has negative charge on the wire side of the plate and positive charge on other side. Field 2 and Field 4 cancel each other out so the the net Efield in the positive plate of CAP 2 and wire connection too supply is zero.

Before I go any further. Let me know if this fits your definition of what's going on. If it does than I have further questions for you.

I made a statement earlier which I believe describes what is happening...

Before equilibrium is reached the negative plate of CAP 1 (connected to the negative plate of the battery) forces electrons to leave the positive plate of CAP 1 and go onto the negative plate of CAP 2

As charge builds up on CAP 2 it increasingly resists this push from CAP 1 until equilibrium is reached and the "push" from CAP 2 equals the "push" from CAP 1.

Here is the diagram again with the two battery terminals added...
BATT ***wire*** CAP 1 ***wire*** CAP 2***wire*** BATT
- ___________ - +(-) ___________ - +(-) __________ +

Lets look at what is happening, but focusing on this region..

CAP1 ***wire*** CAP 2
- +(-) ___________ - +(-)


The negative terminal of the battery creates a "push" via the electric field (at the same time the positive terminal of the battery creates a "pull" via the electric field, but for simplicity I am only looking at the reaction from right to left) and creates a chain reaction through the wire until an electron is added to the negative plate of CAP1.

This electron then "pushes" an electron off of the positive plate of CAP1.
Once again a chain reaction is created in the wire until an extra electron is added to the negative plate of CAP2.

This electron then "pushes" and electron off of the positive plate of CAP2, setting up a chain reaction through the wire until an electron is deposited onto the positive terminal of the battery.

As the number of electrons pushed onto the negative plate of CAP2 increases so does the electric field on that plate. These electrons resist more electrons being added to the plate until finally the "push" on the electrons in the wire from the negative plate of CAP2 is as great as the "push" from the negative plate of CAP1.

At this point the negative plate of CAP1 can't expel any more electrons off of the positive plate of CAP1, because the "push" from the negative plate of CAP2 won't allow it.

So now the best that the negative plate of CAP1 can do is just to sit there and induce a charge separation on the positive plate of CAP1.

No current can flow in between the two capacitors because they are canceling each other out...both "pushing" the electrons in the wire between them in opposite directions with equal force.




One more thing...the electrons are repelled not only by the electrons on the capacitor plates, but also by neighboring electrons in the wire itself.

Maybe picture it like a line of billiard balls all touching each other in a narrow tube where both ends are being pushed on with equal force.

My explanations and analogies are overly simplistic, and the situation is more involved and complex than what I have described, but maybe my thoughts will help you picture this scenerio and allow you to move on to some other mystery to haunt you and keep you up at night.






Russ

 

[meldave00]

Capacitors in Series

Russ,

BATT ***wire*** CAP 1 ***wire*** CAP 2***wire*** BATT
- ___________ - +(-) ___________ - +(-) __________ +

Let me recap your thoughts quickly to make sure that I got this correct.

1). Negative terminal of battery deposits electrons on neg. plate of cap 1

2). Electrons are then repelled from pos. plate of cap 1 and pushed
toward the negative plate of cap 2

3). The increasing concentration of electrons on the neg. plate of cap 2
resist any more electrons to be deposited on the neg. plate of cap2
from pos. plate of cap 1.

4). Equilibrium is established in the region shown below.
CAP1 ***wire*** CAP 2
- +(-) ___________ - +(-)

5). Increasing amounts of electrons deposited on neg. plate of cap 2
repel electrons from pos. plate of cap 2 and push electrons into the
pos. terminal of the battery.

6). Circuit gets put into equilibrium and current flow stops.

That's all I have for now. Let me know if this is correct. Once we are in agreement, I will then give you some more challenging questions about this circuit.

regards,

David

 

[Firefox123]

Russ,

BATT ***wire*** CAP 1 ***wire*** CAP 2***wire*** BATT
- ___________ - +(-) ___________ - +(-) __________ +

Let me recap your thoughts quickly to make sure that I got this correct.

1). Negative terminal of battery deposits electrons on neg. plate of cap 1

2). Electrons are then repelled from pos. plate of cap 1 and pushed
toward the negative plate of cap 2

3). The increasing concentration of electrons on the neg. plate of cap 2
resist any more electrons to be deposited on the neg. plate of cap2
from pos. plate of cap 1.

4). Equilibrium is established in the region shown below.
CAP1 ***wire*** CAP 2
- +(-) ___________ - +(-)

5). Increasing amounts of electrons deposited on neg. plate of cap 2
repel electrons from pos. plate of cap 2 and push electrons into the
pos. terminal of the battery.

6). Circuit gets put into equilibrium and current flow stops.

That's all I have for now. Let me know if this is correct. Once we are in agreement, I will then give you some more challenging questions about this circuit.

regards,

David

That sounds right to me. We are in 100% agreement.

 

[meldave00]

I will reply soon. I have been busy

 

[meldave00]

Russ,

O.K. Now that we are in agreement on what is going on in the circuit. There are some additional questions I have that I don't fully understand. I'm not sure if you know that answers or not either. However, here they are.

1). In the following region. I put numbers in for easier reference.

BATT ***wire*** CAP 1 *****wire***** CAP 2******wire*** BATT
1- ______2______ 3- 4+(5-) ____6______ 7- 8+(9-) ____10_____ 11+

Questions
1). Where is the boundary of 4+ to (5-)? Is it right at the cap to wire boundary or does positive charge extend past the plate and move into the wire. I'm envisioning that positive charge starts to push down the wire with increasing battery voltage.

2). Are electrons evenly distributed from the points of (5-), 6, 7- region? Or is there a gradient from (5-), 6, 7-? If there is a gradient, I'm assuming that maximum amount of electrons are at 7- and the least amount is at (5-).

3). On all wires in the circuit, I'm assuming that all charge reside on the surface of the wire. Do you believe this to be true? Is this also true for the electrons in the (5-), 6, 7- region?

4). For the Q = CV equation. Is Q just the electrons that reside on the capacitors plates or does it include charge that reside in the wires as well. I'm expecially interested if this is the case in the (5-), 6, 7- region?


Sorry I have more questions than answers. Let's stop here and see what you have to say.

regards,

David

 

[Firefox123]

Hello again David...

Russ,

O.K. Now that we are in agreement on what is going on in the circuit. There are some additional questions I have that I don't fully understand. I'm not sure if you know that answers or not either. However, here they are.

1). In the following region. I put numbers in for easier reference.

BATT ***wire*** CAP 1 *****wire***** CAP 2******wire*** BATT
1- ______2______ 3- 4+(5-) ____6______ 7- 8+(9-) ____10_____ 11+

Okay...

1). Where is the boundary of 4+ to (5-)? Is it right at the cap to wire boundary or does positive charge extend past the plate and move into the wire. I'm envisioning that positive charge starts to push down the wire with increasing battery voltage.

I would say that all of the electrons (and lack thereof) are all physically on the plate itself in this interface. I doubt that we ever have a case where the "lack" of electrons extends into the wire. (think about how many electrons are on thsoe capacitor plates!)

2). Are electrons evenly distributed from the points of (5-), 6, 7- region? Or is there a gradient from (5-), 6, 7-? If there is a gradient, I'm assuming that maximum amount of electrons are at 7- and the least amount is at (5-).

I thought there might be a gradient at first also, but after further thought I believe the electrons in the wire in between distribute themselves basically how they would on any conductor in electrostatic equilibrium. They try to get as far from each other as possible and are basically "evenly" distributed.

3). On all wires in the circuit, I'm assuming that all charge reside on the surface of the wire. Do you believe this to be true? Is this also true for the electrons in the (5-), 6, 7- region?

The actual electrons that are part of the wire itself are actually throughout the wire, not just on the surface...remember the rule is that any excess charge placed on a conductor in static equilibrium resides on the surface of the conductor.

"Excess" here means charge not associated with the material itself.

4). For the Q = CV equation. Is Q just the electrons that reside on the capacitors plates or does it include charge that reside in the wires as well. I'm expecially interested if this is the case in the (5-), 6, 7- region?

Q=CV is referring to the charges on the capacitor only.

Sorry I have more questions than answers. Let's stop here and see what you have to say.

No problem.



Russ

 

[meldave00]

Russ,

Sorry I have not been able to get on here as of late. But if you don't mind I would like to bombard you with a few more questions.

Again, I am inserting our text capacitor circuit for reference.

BATT ***wire*** CAP 1 *****wire***** CAP 2******wire*** BATT
1- ______2______ 3- 4+(5-) ____6______ 7- 8+(9-) ____10_____ 11+

Question 1.
Pretty much all physics books and electronic books say the same thing for capacitors in series. Assume Q1neg = Q1pos = Q2neg = Q2pos. Where Q's are referring to the number of charge. First, do you agree with what the books are saying? Second, can you comment on exactly what areas of our text capacitor circuit this is implying? I haven't forgotten about your explanation of capacitors in series, however I am now trying to link your thoughts to what the textbooks are saying.

Question 2.
Do you think that it is true that the battery can only expel an electron from its negative terminal if an electron is sucked into its positive terminal. Than is, no electron can leave the battery without some available to enter?

Question 3.
Let's imagine this circuit. Single capacitor circuit.
BATTneg ***wire*** CAP 1 ******wire*** BATTpos.
If you care to, please try to describe all the Efields in the circuit when the circuit is in equilibrium. From what I understand, the only Net Efield exists from the positive plate to the negative plate of the capacitor. However, there are additional Efields extending out into the wires and such that are canceled out. The attached picture (click to enlarge) shows the Efield of two plates that have positive and negative charge respectively. It shows two Efields going out from both sides of the positively charged plate and shows two Efield coming in from both sides of the negatively charged plate. If shows that on the outsides of each plate that the positive and negative Efields cancel each other, but do not cancel each other between the plates. This to me makes sense. However, this is the case where an opposite Efield from one plate extends passed the outer side of another plate and cancels out the plate to wire Efield of the plate it extended passed. My question is... what happens if we stick a battery across these plates that introduces it own Efield

regards,

David

 

[Firefox123]

Hi david...

Just wanted to let you know that I did see your last post and I am planning on answering it tomorrow if time permits.

Till then.



Russ

 

[Firefox123]

Russ,

Sorry I have not been able to get on here as of late. But if you don't mind I would like to bombard you with a few more questions.

Hi David. Bomb away.

Again, I am inserting our text capacitor circuit for reference.

BATT ***wire*** CAP 1 *****wire***** CAP 2******wire*** BATT
1- ______2______ 3- 4+(5-) ____6______ 7- 8+(9-) ____10_____ 11+

Question 1.
Pretty much all physics books and electronic books say the same thing for capacitors in series. Assume Q1neg = Q1pos = Q2neg = Q2pos. Where Q's are referring to the number of charge. First, do you agree with what the books are saying? Second, can you comment on exactly what areas of our text capacitor circuit this is implying? I haven't forgotten about your explanation of capacitors in series, however I am now trying to link your thoughts to what the textbooks are saying.

Yes. I do agree that the 'Q's are the number of charges on the plates.

As far as the areas go let's look at areas 3,4,5 and 7,8,9. Since these are basically the same setup what we can conclude about 3,4,5 applies to 7,8,9.

Imagine an electron is removed from + plate (region 4) and that same electron is then placed on - plate (region 3). Where will this extra electron reside (I specifically say "extra" here because this electron is not a part of the orignal electrons that make up the negative metal plate, it was added to - plate {region 3} and removed from + plate {region 4}) This electron will reside on the outer surface of the - plate.

On the + plate we are missing an electron so there is a "net positive" voltage. This happens for many electrons on the plates.

Even when we are in an electrostatic equilibrium and no current is flowing the negative charge from the - plate still repels electrons from the + plate. So we get a separation of charge on the + plate. Keep in mind though that the + plate is still positively charged overall despite any charge seperation.

Question 2.
Do you think that it is true that the battery can only expel an electron from its negative terminal if an electron is sucked into its positive terminal. Than is, no electron can leave the battery without some available to enter?

Im not sure to be honest...Im sure we could imagine a hypothetical situation where an electron could leave the negative terminal and not get replaced. But hypotheticals aside, a battery has internal chemistry that is maintaining the voltage across the terminals and is allowing an electron to "leave" the negative terminal while the positive terminal "accepts" an electron. You might want to read up on how batteries work.

Question 3.
Let's imagine this circuit. Single capacitor circuit.
BATTneg ***wire*** CAP 1 ******wire*** BATTpos.
If you care to, please try to describe all the Efields in the circuit when the circuit is in equilibrium. From what I understand, the only Net Efield exists from the positive plate to the negative plate of the capacitor. However, there are additional Efields extending out into the wires and such that are canceled out. The attached picture (click to enlarge) shows the Efield of two plates that have positive and negative charge respectively. It shows two Efields going out from both sides of the positively charged plate and shows two Efield coming in from both sides of the negatively charged plate. If shows that on the outsides of each plate that the positive and negative Efields cancel each other, but do not cancel each other between the plates. This to me makes sense. However, this is the case where an opposite Efield from one plate extends passed the outer side of another plate and cancels out the plate to wire Efield of the plate it extended passed. My question is... what happens if we stick a battery across these plates that introduces it own Efield
View attachment 7342

Could you give me the site where you got that diagram from? Id like to see how they are describing the figure..

Anyway...I think I understand what the E-fields in the figure are and it looks okay to me.


Okay for a single capacitor circuit with a battery connected to it.

A battery is kind of like a capacitor in a way, just that a battery has interal chemistry that maintains the voltage across it while a capacitor does not.

In equilibrium we have an electric field across the battery terminals and an electric field across the capacitor plates. Those two electric fields cancel each other out such that neither the capacitor nor the battery can move any electrons. They are both pushing electrons with equal force, but in opposite directions, within the wire so nothing moves.


Does that make sense?




Russ

 

[meldave00]

Russ,

Thanks for the response. I will respond later when today when I have more time.

David